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Calculate conditional probabilities using formulas and two-way tables.
Learn step-by-step with practice exercises built right in.
Conditional probability: Probability of given has occurred
A two-way table shows 200 students: 120 took calculus, 80 took statistics. 50 took both. Find P(Calculus | Statistics).
P(Calculus | Statistics) = P(Calculus ∩ Statistics) / P(Statistics) = 50/80 = 5/8 = 0.625. Of the 80 students taking statistics, 50 also took calculus, so the conditional probability is 62.5%.
A medical test has sensitivity 0.95 (P(positive | disease)) and specificity 0.90 (P(negative | no disease)). If disease prevalence is 0.02, find P(disease | positive) using Bayes' theorem.
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Intuition: Restrict sample space to outcomes where occurred; within that, find proportion where also occurred
Notation: "" reads as "probability of given "
Data: 200 AP Statistics students
| Passed Exam | Failed Exam | Total | |
|---|---|---|---|
| Attended Class | 95 | 5 | 100 |
| Skipped Class | 40 | 60 | 100 |
| Total | 135 | 65 | 200 |
Question: What's the probability a student passed, given they attended class?
Question: What's the probability a student attended class, given they passed?
Note:
For sequential events:
Start
├─ A (prob=0.4)
│ ├─ C (prob=0.7)
│ └─ D (prob=0.3)
└─ B (prob=0.6)
├─ C (prob=0.2)
└─ D (prob=0.8)
Path probabilities: Multiply along branches
Total:
Or equivalently:
Use: To find probability of both events
"Given that" in problem? → Use conditional probability Computing first? → Use multiplication rule
Always redraw two-way table or tree for conditional probability problems. Shading/circling the condition helps avoid errors.
Let D = disease, + = positive test. P(D|+) = P(+|D)P(D) / [P(+|D)P(D) + P(+|¬D)P(¬D)]. P(+|¬D) = 1 - 0.90 = 0.10. P(D|+) = (0.95 × 0.02) / [(0.95 × 0.02) + (0.10 × 0.98)] = 0.019 / 0.117 ≈ 0.162. Only about 16.2% of positive tests indicate disease; this shows the importance of base rates in medical testing.
In a deck of 52 cards, what is P(second card is an ace | first card was an ace, drawn without replacement)?
After removing one ace, 3 aces remain out of 51 cards total. P(2nd ace | 1st ace) = 3/51 = 1/17 ≈ 0.0588. Without replacement, the sample space shrinks, lowering the conditional probability of a second ace.